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Math Help - Another Probability.

  1. #1
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    Another Probability.

    1) Four friends, 2 females and 2 males are playing contract bridge. Partners are randomly assigned. What is the probability that two females will be partners for the first game?

    I did:
    P(2 females) = P(no 2 females)
    = P(6C2)
    (8C2)
    = 15
    28

    and i got it wrong....dont know what i did wrong.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Just list the simple events, there aren't that many of them.
    Last edited by matheagle; March 25th 2010 at 03:24 PM.
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  3. #3
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    Hello, questiongirl!

    This is easier/harder than we'd expect . . .


    Four friends, 2 females and 2 males are playing contract bridge.
    Partners are randomly assigned.
    What is the probability that two females will be partners for the first game?

    Suppose the four friends are: . \{F_1,\:F_2,\:M_1,\:M_2\}


    There are only three assignments for partners:

    . . \{F_1,F_2\:|\:M_1,M_2\}\quad \{F_1,M_1\:|\:F_2,\,M_2\} \quad \{F_1,\,M_2\:|\:F_2\,M_1\}

    Therefore: . P(F_1,F_2) \;=\;\frac{1}{3}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Choose a female.
    . . Either one, it makes no difference.

    Now what is the probability that her partner is the other female?

    There are 3 choices: . \text{other F},\;M_1,\;M_2

    Therefore: . P(F_1F_2) \;=\;\frac{1}{3}

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  4. #4
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    Thanks Soroban...you see i have misinterpret the question. i thought it means that there are 4 friends, 2females, 2males so in total their are 8people.
    okay i get it. thank you!
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